Question

A ladder 25m long reaches a window of a building 20m above the ground. Determine the distance of foot of ladder from the building.​

Answer 1

$\bigstar\bf\underline{\underline{Given:-}}$

• A ladder "25 m" long
• Reaches a window of a building "20 m" about the ground

$\bigstar\bf\underline{\underline{To \ find:-}}$

• The distance of foot of ladder from the building.

$\bigstar\bf\underline{\underline{Answer:-}}$

Let AC be the ladder and A be the position of the window

Then,

AC = 25 m , AB = 20 m

By using Pythagoras theorem,

∴ AB² = AC² + BC²

$\implies\displaystyle{(25 m)^2 = (20 m)^2 + BC^2}$

$\implies\displaystyle{BC^2 = 625 m^2 - 400 m^2}$

$\implies\displaystyle{BC^2 = 225 m^2}$

$\implies\displaystyle{BC^2 = (15 m)^2}$

(Here, both square's gets cancelled)

$\implies\boxed{\boxed{\underline{\bf\red{BC = 15 cm}}}}$

Therefore, the distance of the foot of ladder from the building is "15 cm"

Answer 2

★Given :

• Length of ladder = 25cm
• It reaches a window of a building 20m above the ground.

★To find :

• The distance of the foot of the ladder from the building.​

★Solution :

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf 20}\put(2.8,.3){\large\bf 15}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf \theta }\end{picture}$

We are given the length of the ladder and the length of the building and are supposed to find  the distance of the foot of the ladder from the building i.e. BC. From the figure ,

• AB = 20cm (perpendicular)
• AC = 25m (hypotenuse)
• BC = ? (Base)

Using the Pythagoras theroem,

$\leadsto \underline{\boxed{\bold{Hypotenuse^2 = Perpendicular^2+Base^2 }}}$

Substituting the values,

$\sf \implies AC^2 = AB^2 + BC^2 \\ \\ \sf \implies BC^2 = AC^2 - AB^2 \\\\\sf \implies BC^2 = (25)^2 - (20)^2 \\\\\sf \implies BC^2 = 625 - 400 \\\\\sf \implies BC^2 = 225 \\\\\implies \sf BC = \sqrt{225} \\\\\implies \underline{\boxed{\bold{BC = 15m}}}$

∴ The distance of foot of ladder from the building is 15m.

    _______________